📚 AS AQA Mathematics Unit 1 (Jan 2022) Paper Walkthrough | AQA AS 数学 单元1（2022年1月）试卷精讲
This article gives a step-by-step walkthrough of the January 2022 AQA AS Mathematics Unit 1 paper. The focus is on high-mark questions and common pitfalls, so that you can convert your revision time into exam marks.
本文逐题精讲2022年1月AQA AS数学单元1试卷。我们聚焦高分题及常见失分点，帮助你把复习时间转化为考卷上的分数。
1. Expanding and Factorising Quadratics | 展开与因式分解二次式
so split the middle term: 6x² − 15x + 4x − 10 = 3x(2x − 5) + 2(2x − 5) = (3x + 2)(2x − 5).
因式分解 6x² − 11x − 10。找到两个数，乘积等于 6 × −10 = −60，和为 −11。这两个数是 −15 和 4，因此拆中项：6x² − 15x + 4x − 10 = 3x(2x − 5) + 2(2x − 5) = (3x + 2)(2x − 5)。
Always multiply your brackets back out to check the expansion.
务必将括号重新乘开，以检验因式分解是否正确。
2. Equation of a Straight Line | 直线方程
s (9 − 5) / (4 − 2) = 2. Using y − y₁ = m(x − x₁), we get y − 5 = 2(x − 2), so y = 2x + 1.
直线过点 (2, 5) 和 (4, 9)。斜率为 (9 − 5) / (4 − 2) = 2。由点斜式 y − y₁ = m(x − x₁)，得 y − 5 = 2(x − 2)，即 y = 2x + 1。
Check your answer by substituting both given points into your final equation.
将两个已知点分别代入最终方程进行验证。
3. Differentiation from First Principles | 用第一原理求导
ting f(x) leaves 2xh + h², and dividing by h gives 2x + h. As h → 0, the derivative is 2x.
对于 f(x) = x²，使用定义式 f ′(x) = lim(h→0) [ f(x + h) − f(x) ] / h。则 f(x + h) = (x + h)² = x² + 2xh + h²。减去 f(x) 后剩 2xh + h²，除以 h 得 2x + h。当 h → 0 时，导数为 2x。
In the exam, write down each step of the limit process to earn method marks.
考试中请写出极限过程的每一步，以获得方法分。
4. Binomial Expansion | 二项式展开
Expand (1 + x/2)⁴ up to and including the term in x². The coefficients from Pascal' + x/2)⁴ = 1 + 4(x/2) + 6(x/2)² + 4(x/2)³ + (x/2)⁴ = 1 + 2x + (3/2)x² + (1/2)x³ + (1/16)x⁴.
展开 (1 + x/2)⁴，写到含 x² 的项。杨辉三角的系数是 1, 4, 6, 4, 1。因此 + x/2)⁴ = 1 + 4(x/2) + 6(x/2)² + 4(x/2)³ + (x/2)⁴ = 1 + 2x + (3/2)x² + (1/2)x³ + (1/16)x⁴。
If only the x² term is asked for, the first three terms are enough.
若题目只要求到 x² 项，计算前三项即可。
5. Solving a Quadratic Inequality | 二次不等式求解
itive quadratic, so it lies below the x-axis between the roots. Therefore the solution is −1 < x < 6.
解不等式 x² − 5x − 6 < 0。先因式分解：(x − 6)(x + 1) < 0。临界值为 −1 和 6。由于 y = x² − 5x − 6 是开口向上的二次函数，其图像在两个根之间位于 x 轴下方。因此解为 −1 < x < 6。
Be careful with strict inequalities; the endpoints are not included.
注意严格不等号，端点值不应包含在解集内。
6. Integration and Area Under a Curve | 积分与曲线下面积
at the limits: (27 + 18 − 15) − (1 + 2 − 5) = 30 − (−2) = 32. The area is 32 square units.
求曲线 y = 3x² + 4x − 5 在 x = 1 到 x = 3 之间与 x 轴所围成的面积。不定积分为 ∫(3x² + 4x − 5) dx = x³ + 2x² − 5x + C。代入上下限：(27 + 18 − 15) − (1 + 2 − 5) = 30 − (−2) = 32。面积为 32 平方单位。
Sketch the curve first to confirm that the region lies entirely above the x-axis between the limits.
先画草图，确认所给区间内曲线完全位于 x 轴上方。
7. Statistical Measures: Mean and Standard Deviation | 统计量：均值与标准差
/ 6 = 966 / 6 = 161 cm. For the standard deviation, use the formula s = √[ (∑x² / n) − (x̄)² ], where ∑x² = 154² + 158² + 160² + 162² + 165² + 167² = 155638.
六名学生的身高（cm）为 154, 158, 160, 162, 165, 167。平均身高 = (154 + 158 + 160 + 162 + 165 + 167) / 6 = 966 / 6 = 161 cm。计算标准差时使用公式 s = √[ (∑x² / n) − (x̄)² ]，其中 ∑x² = 154² + 158² + 160² + 162² + 165² + 167² = 155638。
Hence s = √155638 / 6 − 161² = √25939.67 − 25921 = √18.67 ≈ 4.32 cm.
因此 s = √155638 / 6 − 161² = √25939.67 − 25921 = √18.67 ≈ 4.32 cm。
8. Probability Tree | 概率树
Two balls are drawn without replacement. The probability that both are red is P(R₁) × P(R₂ | R₁) = (4/10) × (3/9) = 12/90 = 2/15. The probability of one of each colour is P(R ∩ B) + P(B ∩ R) = (4/10 × 6/9) + (6/10 × 4/9) = 24/90 + 24/90 = 48/90 = 8/15.
袋中有4个红球和6个蓝球。不放回地取出两个球。两个都是红球的概率为 P(红₁) × P(红₂ | 红₁) = (4/10) × (3/9) = 12/90 = 2/15。一红一蓝的概率为 P(红∩蓝) + P(蓝∩红) = (4/10 × 6/9) + (6/10 × 4/9) = 24/90 + 24/90 = 48/90 = 8/15。
For "without replacement", the denominator always decreases by one after the first draw.
当题为“不放回”时，第一次抽取后分母必须减一。
9. Stationary Points | 驻点分析
x = 1 or x = 3. The y-values are y(1) = 1 − 6 + 9 + 1 = 5 and y(3) = 27 − 54 + 27 + 1 = 1.
对于曲线 y = x³ − 6x² + 9x + 1，求驻点并判断其性质。求导：dy/dx = 3x² − 12x + 9 = 3(x² − 4x + 3) = 3(x − 1)(x − 3)。令 dy/dx = 0，得 x = 1 或 x = 3。对应的 y 值为 y(1) = 1 − 6 + 9 + 1 = 5，y(3) = 27 − 54 + 27 + 1 = 1。
The second derivative is
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